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What is the name of the following function (if there is one)?

$$f(x) = \begin{cases} x & \text{ if } -1 \leq x \leq 1\\ \frac 1x & \text{ if } x < -1 \text{ or } x >1\\ \end{cases} $$

If this function has a name, how is it usually denoted?

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Perhaps the context where you came across this might help... –  Aryabhata May 25 '11 at 1:34
    
I've never seen the function in any particular context. I was trying to find a function so that for numbers $a$ and $b$ greater than 0: $f(\frac{a}{b})$ is always less than or equal to 1; $f(\frac{a}{b})$ is largest when $a=b$; $f(\frac{a}{b})$ is increasing when b>a; $f(\frac{a}{b})$ is decreasing when a>b. –  objectivesea May 25 '11 at 2:08
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Why would you expect it to have a name? It's equal to $\min(1, x^2)/x$ (except for $x=0$). –  Robert Israel May 25 '11 at 2:11
    
This function seems like it would be useful in many cases. Another way of writing the function: For numbers $a$ and $b$, $f(\frac{a}{b})=\text{min}(\frac{a}{b},\frac{b}{a})$. Isn't it often that a ratio less than one is needed, rather than the ratio's inverse? –  objectivesea May 25 '11 at 2:21
    
$\mathrm{sgn}(x)e^{-|\log(|x|)|}$ ? –  robjohn Jul 8 '12 at 17:03

1 Answer 1

If there doesn't exist a standard name, I'd call it the "left one-right one-identity, alter-inverse" function on the reals. The "left one-right one identity" part means that from the "left one" $(-1)$ to "the right one" the function matches that of the identity function on the reals. The "alter-inverse" means "otherwise we have the inverse" (in Latin alter means "other"), since $1/x$ comes as the inverse function from the reals without 0 to the reals without 0. The function maps to $[-1, 1]$ which under real-number multiplication forms a commutative monoid [$([-1, 1], \ast)$ is the commutative monoid] with annihilator of $0$, just like the reals do. Have you found an inverse for your function?

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